Realism from Locality? Bell's Theorem & Nonlocality in QM

[Demystifier]

I think what's also often missing in all these discussions (including the ones running in the parallel thread on entanglement swaping a la Zeilinger et al) is that never the complete state of the entangled photons is considered. It's just written down the polarization part but not the momentum part, which is however important in the considerations of locality arguments. I think this deserves another FAQ/Insight article. Maybe then my point of view becomes clearer too.

It's in fact straightforward to add the momentum part, but it doesn't help to understand the issue of (non)locality. Usually only the polarization part is analyzed explicitly because only the polarization is measured in experiments that violate Bell-like inequalities.

 

[A. Neumaier]

I think what's also often missing in all these discussions (including the ones running in the parallel thread on entanglement swaping a la Zeilinger et al) is that never the complete state of the entangled photons is considered. It's just written down the polarization part but not the momentum part, which is however important in the considerations of locality arguments. I think this deserves another FAQ/Insight article. Maybe then my point of view becomes clearer too.

The momentum part is trivial; just take for each particle's part the tensor product with a smeared momentum state. Up to a joint phase, these remain constant throughout, hence can be factored out.

 

[vanhees71]

Sure, writing it in the form of $|\text{momentum part} \rangle \otimes |\text{polarization part} \rangle$ it's (here for the singlet Bell state in the polarization part)
$$(|\vec{p}_1\rangle \otimes |\vec{p}_2 \rangle - |\vec{p}_2 \rangle \otimes |\vec{p}_1 \rangle) \otimes (|H \rangle \otimes |V \rangle -|V \rangle \otimes H \rangle).$$
For wave packets just integrate over the momenta with some $L^2$-function weight (e.g., a Gaussian).

 

[zonde]

Sure, writing it in the form of $|\text{momentum part} \rangle \otimes |\text{polarization part} \rangle$ it's (here for the singlet Bell state in the polarization part)
$$(|\vec{p}_1\rangle \otimes |\vec{p}_2 \rangle - |\vec{p}_2 \rangle \otimes |\vec{p}_1 \rangle) \otimes (|H \rangle \otimes |V \rangle -|V \rangle \otimes H \rangle).$$
For wave packets just integrate over the momenta with some $L^2$-function weight (e.g., a Gaussian).

Entanglement is between particles not observables. The state you wrote is just meaningless as polarization part already includes momentum:
$|H\rangle_{\vec{p}_1}\otimes|V\rangle_{\vec{p}_2} -|V\rangle_{\vec{p}_1}\otimes|H\rangle_{\vec{p}_2}$

 

[vanhees71]

Entanglement is a property of states.

What I wrote down is
$$|\vec{p}_1,H \rangle \otimes |\vec{p}_2,V \rangle - |\vec{p_1},V \rangle \otimes |\vec{p}_2,H \rangle - |\vec{p}_2,H \rangle \otimes |\vec{p}_1,V \rangle +|\vec{p}_2,V \rangle \otimes |\vec{p}_1,H \rangle
= [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) - \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2 H)] |\Omega \rangle.$$
The corresponding normalized state ket is given by this vector $\times 1/2$.

It takes into account the Bose nature of photons. Your state is not symmetrized as a two-particle (two-photon) bosonic state should be.

 

[PeterDonis]

polarization part already includes momentum

I have no idea what you mean by this, since the polarization degrees of freedom are different from the momentum degrees of freedom; neither one can "include" the other.

It might be the case that the momentum and polarization degrees of freedom are entangled, but that is not the same as polarization "including" momentum.

 

[zonde]

I have no idea what you mean by this, since the polarization degrees of freedom are different from the momentum degrees of freedom; neither one can "include" the other.

It might be the case that the momentum and polarization degrees of freedom are entangled, but that is not the same as polarization "including" momentum.

Well, maybe "included" is not the right word. Polarization modes are indexed by spatial directions. In entaglement measurements interference happens between polarization modes in the same spatial direction. No interference is created between different spatial directions. That relationship between different spatial directions is revealed only indirectly by comparing outcomes of polarization mode interference measurements.
So it's all tied together. Two polarization modes and two spatial directions makes four combined modes that are measured pairwise and then compared.

 

[vanhees71]

I've no clue what you are talking about. Let's talk about two photons with momenta $\vec{p}_1$ and $\vec{p}_2$. The natural complete one-photon basis is the momentum-helicity-eigenbasis $|\vec{p},\lambda$ where $\vec{p} \in \mathbb{R}^3$ and $\lambda \in \{-1,1\}$. The helicity is the projection of the total angular momentum (sic!) to the direction of momentum.

A basis for the two-photon states are
$$|1_{\vec{p}_1,\lambda_1},1_{\vec{p}_2,\lambda_2} \rangle = \frac{1}{\sqrt{2}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1) \hat{a}^{\dagger}(\vec{p}_2,\lambda_2) |\Omega \rangle.$$
By definition these states describe non-entangled photons. Note however that these are not simply product states of single photons, but the creation operators automatically imply the symmetrization necessary to describe the photons as indistinguishable bosons.

A basis of maximally entangled two-photon states is given by
$$|\phi_{12}^{\pm} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,1) \hat{a}^{\dagger}(\vec{p}_2,1) \pm \hat{a}^{\dagger}(\vec{p}_1,-1) \hat{a}^{\dagger}(\vec{p}_2,-1) \right]$$
and
$$|\psi_{12}^{\pm} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,1) \hat{a}^{\dagger}(\vec{p}_2,-1) \pm \hat{a}^{\dagger}(\vec{p}_1,-1) \hat{a}^{\dagger}(\vec{p}_2,1) \right].$$
For details about entanglement of indistinguishable particles, see e.g.

J. C. Garrison, R. Y. Chiao, Quantum Optics, Oxford University Press (2008)

 

[PeterDonis]

Polarization modes are indexed by spatial directions.

No, they aren't. I think you are confusing photon polarization with spin measurements on spin-1/2 particles like electrons. Both lead to the Hilbert space of a qubit for the relevant degrees of freedom, but the physical measurements are not the same.

 

[zonde]

No, they aren't. I think you are confusing photon polarization with spin measurements on spin-1/2 particles like electrons. Both lead to the Hilbert space of a qubit for the relevant degrees of freedom, but the physical measurements are not the same.

Of course they are. Look here: Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory, page 2
Here spatial directions are called "idler" and "signal". If you have any doubts that "idler" and "signal" are spatial directions in page 3 you can read sentence:
"By placing polarizers rotated to angles $\alpha$ and $\beta$ in the signal and idler paths, respectively, we measure the polarization of the downconverted photons."

 

[PeterDonis]

Here spatial directions are called "idler" and "signal".

Yes, those are spatial directions, which are different degrees of freedom from the polarization degrees of freedom. Different polarization measurements are performed on the photons going in those two spatial directions, but that does not mean the polarization degrees of freedom are the same as the spatial direction degrees of freedom.

 

[zonde]

Yes, those are spatial directions, which are different degrees of freedom from the polarization degrees of freedom. Different polarization measurements are performed on the photons going in those two spatial directions, but that does not mean the polarization degrees of freedom are the same as the spatial direction degrees of freedom.

Of course polarization degrees of freedom and spatial direction degrees of freedom are different. There are both orthogonal polarization modes in each spatial direction for entangled state. And that's what goes into description of entangled state. There is $|H_s\rangle$ and there is $|H_i\rangle$. So when someone writes entangled state omitting these subscripts, like: $|\psi^+\rangle=\frac{1}{\sqrt{2}}(|H\rangle\otimes|V\rangle\ + |V\rangle\otimes|H\rangle)$ different spatial directions are implied for the same polarization modes.

 

[zonde]

I've no clue what you are talking about.

You can take a look at this paper: Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory

 

[vanhees71]

Well, this paper uses the usual simplification, treating the photons as if they were distinguishable. The correct polarization-singlet state, idealized as momentum eigenstates is
$$|\Psi^{-} \rangle = \frac{1}{2} [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) -\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) ] |\Omega \rangle.$$
The location of the registration events of these photons comes in by placing the detectors at their locations and then what's well defined are the two-photon-detection and/or single-photon probabilities given by the expectation values of appropriate electric-field operators.

 

[zonde]

Well, this paper uses the usual simplification, treating the photons as if they were distinguishable. The correct polarization-singlet state, idealized as momentum eigenstates is
$$|\Psi^{-} \rangle = \frac{1}{2} [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) -\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) ] |\Omega \rangle.$$

Does this "correct" state description produce the same predictions for entanglement measurements as the "incorrect" one?

 

[zonde]

There is an experiment demonstrating polarization entangled photons with different wavelength:
Three-color Sagnac source of polarization-entangled photon pairs
It seems that polarization entangled photons can be quite distinguishable.

 

[PeterDonis]

when someone writes entangled state omitting these subscripts, like: $|\psi^+\rangle=\frac{1}{\sqrt{2}}(|H\rangle\otimes|V\rangle\ + |V\rangle\otimes|H\rangle)$ different spatial directions are implied for the same polarization modes.

Different spatial directions are implied for the first ket vs. the second ket in each term. That's because each ket (first and second in each term) refers to a distinct photon, which is distinguishable by the direction in which it is moving. The more complete expressions that @vanhees71 is writing down make all this explicit.

I don't know if this is what you mean by "the same polarization modes" or not, but if it is, your choice of words is very poor, and your posts are only going to cause confusion.

 

[vanhees71]

Does this "correct" state description produce the same predictions for entanglement measurements as the "incorrect" one?

Fortunately yes.

 

[vanhees71]

There is an experiment demonstrating polarization entangled photons with different wavelength:
Three-color Sagnac source of polarization-entangled photon pairs
It seems that polarization entangled photons can be quite distinguishable.

Well, what I wrote is a general two-photon state with different momenta. Of course you can also have $E_j=\hbar \omega_j=c |\vec{p}_j|$ different.

Nevertheless the photons are indistinguishable in the sense that it's a bosonic and not a product state. Already
$$
|1_{\vec{p}_1,\lambda_1},1_{\vec{p}_2,\lambda_2} \rangle
=\frac{1}{\sqrt{2}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1)
\hat{a}^{\dagger}(\vec{p}_2,\lambda_2) |\Omega \rangle =
\frac{1}{\sqrt{2}} (|\vec{p}_1,\lambda_1 \rangle \otimes
|\vec{p}_2,\vec{\lambda}_2 \rangle + |\vec{p}_2,\lambda_2 \rangle
\otimes |\vec{p}_1,\lambda_1 \rangle)$$
is not a product state.

Of course the photons are distinguishable in some sense by their momenta, and what's indeed entangled for this state is that in this state the photon with momentum $\vec{p}_1$ necessarily carries helicity $\lambda_1$ and that with $\vec{p}_2$ carries $\lambda_2$.

The "different spatial directions" discussed in other postings are of course the different directions of the photons' momenta.

 

[zonde]

Nevertheless the photons are indistinguishable in the sense that it's a bosonic and not a product state. Already
$$
|1_{\vec{p}_1,\lambda_1},1_{\vec{p}_2,\lambda_2} \rangle
=\frac{1}{\sqrt{2}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1)
\hat{a}^{\dagger}(\vec{p}_2,\lambda_2) |\Omega \rangle =
\frac{1}{\sqrt{2}} (|\vec{p}_1,\lambda_1 \rangle \otimes
|\vec{p}_2,\vec{\lambda}_2 \rangle + |\vec{p}_2,\lambda_2 \rangle
\otimes |\vec{p}_1,\lambda_1 \rangle)$$
is not a product state.

I don't follow you. Twin photons generated in PDC process with known output polarizations are not entangled. From what you wrote it seems you claim that such photons are entangled. So I doubt that I understand you correctly.

Of course the photons are distinguishable in some sense by their momenta, and what's indeed entangled for this state is that in this state the photon with momentum $\vec{p}_1$ necessarily carries helicity $\lambda_1$ and that with $\vec{p}_2$ carries $\lambda_2$.

I'm not sure why you call such relationship an "entanglement". Can you use such relationship to violate Bell inequality? I don't see how. To me it seems just like classical determinism.

 

[DrChinese]

I don't follow you. Twin photons generated in PDC process with known output polarizations are not entangled. From what you wrote it seems you claim that such photons are entangled. So I doubt that I understand you correctly.

Not sure if this is the sticking point between you and PeterDonis or not. Generally PDC can EITHER produce polarization entangled pairs OR non-polarization entangled pairs. It depends on the specific setup. I think this is what you are referring to.

 

[vanhees71]

What we are talking about are pdc pairs prepared in one of the four Bell states
$$|\phi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,H) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,V) \right], \\
|\psi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,H) \right].$$
Of course I've idealized this a bit. In reality there are always proper normalizzble Hilbert-space vectors with the momenta having a distribution of finite width.

 

[zonde]

What we are talking about are pdc pairs prepared in one of the four Bell states
$$|\phi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,H) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,V) \right], \\
|\psi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,H) \right].$$
Of course I've idealized this a bit. In reality there are always proper normalizzble Hilbert-space vectors with the momenta having a distribution of finite width.

You can make careful arrangements of PDC crystal(-s) and pump beam so that two different downconversion processes contribute orthogonal polarization modes (or probability amplitudes in theoreticians language) to the same spatial modes. And in addition you can make very careful arrangements so that two polarization modes in the same spatial mode can interfere with very high visibility.
If you do these things you will get polarization entangled photon pairs.

On the other hand if you do only basic arrangements of PDC process you will get photon pairs with perfectly known polarizations. They will still show a lot of correlations, but all these can be explained using correlated LHVs that are determined at the source.
In post #55 I was talking about this second ("raw") type of PDC process.

 

[vanhees71]

What does "LHVs" mean? Of course, one has to do the right manipulations with the photons to get (approximately) the said Bell states. As a theorist (and not a quantum-optics expert) I'm only interested in the fact that the experimentalists can do that somehow with great accuracy.

 

[Demystifier]

What does "LHVs" mean?

Local hidden variables.

 

[zonde]

Of course, one has to do the right manipulations with the photons to get (approximately) the said Bell states. As a theorist (and not a quantum-optics expert) I'm only interested in the fact that the experimentalists can do that somehow with great accuracy.

Of course. It's just sometimes people focus on entangled photons so much that they start to perceive them as the only "natural state" for twin photons. So I wanted to bring up product state twin photons just to be sure that we are on the same page.

I tried to think over your claim in post #54. I'm not sure I understand your (and probably not only your) view about symmetrization. I have seen the explanations about swapping the particles and getting the same situation. Well if we include position into particle description then we are not swapping anything physical, we just swap meaningless labels in representation. But swapping one representation with other should not have any physical consequences.
And then it seems that in QFT we get very elegant treatment of this situation because in QFT particles don't have arbitrary labels. There particle state is either occupied or not so there is nothing to swap.

On the other hand there is absolutely real experimentally observable phenomena called Hong-Ou-Mandel interference which shows that there is something interesting about identical (indistinguishable) photons.

So I view symmetrization as property of dynamical process where two indistinguishable photons at initial states 1 and 2 end up in final states 3 and 4. And the physical property of symmetrization is that this process can not be described as (photon from state 1 transitions to state 3 and photon from state 2 transitions to state 4) OR (photon from state 1 transitions to state 4 and photon from state 2 transitions to state 3).
If I remember correctly I got this explanation form Feynman's "QED: The Strange Theory of Light and Matter".
And this explanation seems to be consistent with HOM interference.

 

[Jimster41]

No, it would be nonlocal and contextual.

Gotta (try to) read the papers. I'm confused here - what constrains context a priori? Is something like all "previous" contexts, i.e. all GR consistent causal world-lines?

[Edit] Pretty interesting. I wish I understood the little proof of theorem 4.1-3 on the "Non existence of non-contextual value-maps". It's obviously key and it's pure math and... I don't quite follow. And I tried to read up on Kochen-Specker... which I also... don't quite follow - but which seems more physical.

To me they sort of tie the idea of the Bhomian Pilot wave to Quantum Gravity Tensor Network Models

 

[vanhees71]

Of course. It's just sometimes people focus on entangled photons so much that they start to perceive them as the only "natural state" for twin photons. So I wanted to bring up product state twin photons just to be sure that we are on the same page.

I tried to think over your claim in post #54. I'm not sure I understand your (and probably not only your) view about symmetrization. I have seen the explanations about swapping the particles and getting the same situation. Well if we include position into particle description then we are not swapping anything physical, we just swap meaningless labels in representation. But swapping one representation with other should not have any physical consequences.
And then it seems that in QFT we get very elegant treatment of this situation because in QFT particles don't have arbitrary labels. There particle state is either occupied or not so there is nothing to swap.

On the other hand there is absolutely real experimentally observable phenomena called Hong-Ou-Mandel interference which shows that there is something interesting about identical (indistinguishable) photons.

So I view symmetrization as property of dynamical process where two indistinguishable photons at initial states 1 and 2 end up in final states 3 and 4. And the physical property of symmetrization is that this process can not be described as (photon from state 1 transitions to state 3 and photon from state 2 transitions to state 4) OR (photon from state 1 transitions to state 4 and photon from state 2 transitions to state 3).
If I remember correctly I got this explanation form Feynman's "QED: The Strange Theory of Light and Matter".
And this explanation seems to be consistent with HOM interference.

My point is much simpler. Photons are indistinguishable bosons. A convenient (generalized) basis for free photons is the momentum-helicity or momentum-linear-polarization-state basis (you cannot easily talk about position representations for photons since photons do not have position observables to begin with). I denote them with $|\vec{p},\lambda \rangle$, where $\lambda \in \{-1,1 \}$ is the helicity of the photon (the projection of the total angular-momentum on the direction of $\vec{p}$). This refers to plane-wave left- and right-circular polarized em. waves.

Now photons are bosons, and thus the correct Hilbert space is a bosonic Fock space, i.e., for each $(\vec{p},\lambda)$ there's an annihilation operator $\hat{a}(\vec{p},\lambda)$ which obeys the bosonic commutation relations $[\hat{a}^{\dagger}(\vec{p},\lambda),\hat{a}^{\dagger}(\vec{p}',\lambda')]=\delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda \lambda'})$. A complete set of (generalized) orthonormal basis vectors are common eigenstates of the occupation-number observables $\hat{N}(\vec{p},\lambda)=\hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda)$,
$$|\{n(\vec{p},\lambda ) \}_{\vec{p} \in \mathbb{R}^3,\lambda \in \{-1,1\}} \rangle = \prod_{\vec{p},\lambda} \frac{1}{\sqrt{n(\vec{p},\lambda)!}} [\hat{a}^{\dagger}(\vec{p},\lambda)^{n(\vec{p},\lambda)}|\Omega \rangle,$$
where $|\Omega \rangle$ is the vacuum (ground) state, for which all $n(\vec{p},\lambda)=0$.

An $n$-photon Fock state is thus from the totally symmetrized part of $n$-fold Kronecker product of the single-photon Hilbert space.

E.g., if for a photon pair you want the polarization-singlet state $|1,-1 \rangle-|-1,1 \rangle$, the spatial (momentum) part must also be antisymmetric. I.e., written in the product basis you have
$$|\psi^{-} \rangle=\frac{1}{2} (|\vec{p}_1 \rangle \otimes |\vec{p}_2 \rangle-|\vec{p}_2 \rangle \vec{p}_1)(|1 \rangle \otimes |-1 \rangle)-|-1 \rangle \otimes |1 \rangle.$$
Multiply this out and you get the completely symmetrized state in the tensor-product notation. This is very cumbersome. With the creation operators it's much more convenient
$$|\psi^- \rangle = \frac{1}{2} (\hat{a}^{\dagger}(\vec{p}_1,1) \hat{a}^{\dagger}(\vec{p}_2,-1) - \hat{a}^{\dagger}(\vec{p}_1,-1) \hat{a}^{\dagger}(\vec{p}_2,1 ) \Omega \rangle.$$
The HOM (Hong-Ou-Mandel) effect in its most simple form can be described as follows. Two photons of the same frequency enter a (symmetric) beam splitter. Let's look at plane-wave modes. Let the incoming two photons be created by $\hat{a}_{\lambda_1}^{\dagger}$ and $\hat{b}_{\lambda_2}^{\dagger}$ respectively. Here we use the abbreviation
$$\hat{a}_{\lambda_1}=\hat{a}(\vec{p}_a,\lambda_1), \quad \hat{b}_{\lambda_2} = \hat{a}(\vec{p}_b,\lambda_2).$$
The incoming two-photon state is
$$|\psi_0 \rangle=\hat{a}_{\lambda_1}^{\dagger} \hat{b}_{\lambda_2}^{\dagger}.$$
The "scattering matrix" describing the balanced symmetric beam splitter is given by the unitary matrix
$$U_{\text{BS}}=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}.$$
This transforms the creation operators of the incoming to those of the outgoing photons
$$\begin{pmatrix} \hat{a}_{\lambda}^{\prime \dagger} \\ \hat{b}_{\lambda}^{\prime \dagger} \end{pmatrix}= U_{\text{BS}} \begin{pmatrix} \hat{a}_{\lambda}^{\dagger} \\ \hat{b}_{\lambda}^{\dagger} \end{pmatrix}=\frac{1}{\sqrt{2}} \begin{pmatrix} \hat{a}_{\lambda}^{\dagger} + \mathrm{i} \hat{b}_{\lambda}^{\dagger} \\ \mathrm{i} \hat{a}_{\lambda}^{\dagger}+\hat{b}_{\lambda}^{\dagger} \end{pmatrix}.$$
Thus after the beam splitter the state is
$$|\psi_2 \rangle=\hat{a}_{\lambda_1}^{\prime \dagger} \hat{b}_{\lambda_2}^{\prime \dagger}|\Omega \rangle = \frac{1}{2} (\hat{a}_{\lambda_1}^{\dagger}+\mathrm{i} \hat{b}_{\lambda_1}^{\dagger})(\mathrm{i} \hat{a}_{\lambda_2}^{\dagger} + \hat{b}_{\lambda_2}^{\dagger})|\Omega \rangle=\frac{\mathrm{i}}{2} (\mathrm{i} \hat{a}_{\lambda_1}^{\dagger} \hat{a}_{\lambda_2}^{\dagger}+\mathrm{i} \hat{b}_{\lambda_1}^{\dagger} \hat{b}_{\lambda_2}^{\dagger} - \hat{a}_{\lambda_2}^{\dagger} \hat{b}_{\lambda_1}^{\dagger}+\hat{a}_{\lambda_1}^{\dagger} \hat{b}_{\lambda_2}^{\dagger})|\Omega \rangle .$$
Now if the photons have different parity, i.e., are distinguishable, you get with probability 1/2 both photons in the same mode (either a or b) and with probability 1/2 in different modes (one in a and one in b). This is as expected from the classical case.

If, however $\lambda_1=\lambda_2=\lambda$ the output state is
$$|\psi_2 \rangle=\frac{1}{\sqrt{2}}(|2_{a,\lambda} \rangle + |2_{b,\lambda}),$$
i.e., both photons always end up in the same mode (with probability 1/2 in a and with probability 1/2 in b).

This is a specific quantum effect of indistinguishable photons and cannot explained in the classical wave picture.

 

[Heikki Tuuri]

If I am the only observer who can make a wave function to collapse, and all other humans are "Wigner's friends", or Schrödinger's cats, does that mean that quantum mechanics is "local" in some sense meant by the original poster DeMystifier?

I support the Many Worlds interpretation where the wave function never collapses. We may, for example, calculate the development of the wave function using Schrödinger's equation. The calculation is "local" in the sense that a computer can calculate the next time step of the wave function using only "local" data for each calculation grid point.

I, as an observing subject, feel that I am now located in a certain branch of the Many Worlds. I, subjectively, have seen the wave function of the universe to "collapse" as this branch. But the whole network of the Many Worlds exists. I think one can say that this interpretation is "local".

 

[PeterDonis]

The calculation is "local" in the sense that a computer can calculate the next time step of the wave function using only "local" data for each calculation grid point.

No, it can't, because the potential term in the Hamiltonian is not local. For example, in the case of a multi-particle system, the potential is a function of the positions of all the particles, which is a nonlocal function.

 

[Heikki Tuuri]

To overcome the problem brought up by member Peter Donis, we need to model all interactions as couplings between a particle and a field, or between two fields.

For example, an electron responds to the electric field at its location, and does not care where the particles who create this field are located. Energy and momentum can be stored in the field.

We know that the self-force between an electron and its field is poorly understood in classical electromagnetism. Thus, the locality is not an empirical fact but a conjecture. We have to conjecture that a consistent local theory of an electron and its field can be formulated.

 

[akvadrako]

No, it can't, because the potential term in the Hamiltonian is not local. For example, in the case of a multi-particle system, the potential is a function of the positions of all the particles, which is a nonlocal function.

Just because there is a non-local method to calculate "the next step" doesn't mean there isn't also a local method. So far nobody has shown an example of an experiment where MWI seems to require non-local information and it certainly hasn't been proven.

 

[A. Neumaier]

We have to conjecture that a consistent local theory of an electron and its field can be formulated.

We have a (at the practical level) consistent local theory of the electron field and the electromagnetic field. It is called QED. The electron is only an approximate, asymptotic concept in this theory.

 

[PeterDonis]

So far nobody has shown an example of an experiment where MWI seems to require non-local information

As an interpretation of QM, the MWI makes all of the same experimental predictions as all other interpretations of QM, including predictions of violations of the Bell inequalities, which is the standard thing being referred to by the term "nonlocality".

 

[akvadrako]

As an interpretation of QM, the MWI makes all of the same experimental predictions as all other interpretations of QM, including predictions of violations of the Bell inequalities, which is the standard thing being referred to by the term "nonlocality".

Well, there is of course "Bell non-locality", but that doesn't apply to the description mentioned by @Heikki Tuuri. Bell non-locality doesn't mean that you can't calculate the future state using only local information.